course Mth 163 Assignment 7 q_a_ and query. ?{????m??`???\?Q?assignment #007007.
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18:14:51 `q001. Note that this assignment has 8 questions Sketch a graph of the following (x, y) points: (1,2), (3, 5), (6, 6). Then sketch the straight line which appears to come as close as possible, on the average, to the four points. Your straight line should not actually pass through any of the given points. Describe how your straight line lies in relation to the points. Give the coordinates of the point at which your straight line passes through the y axes, and give the coordinates of the x = 2 and x = 7 points on your straight line. Determine the slope of the straight line between the last two points you gave.
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RESPONSE --> My line lies about 1/4 of a unit above the point (1 , 2) , about 1/2 of a unit below the point (3 , 5) , and about 1/4 of a unit above the point (6 , 6) It passes through the y-axis at about y-value 1.75 The coordinates of x=2 is (2 , 2.5) The coordinates of x=7 is (7 , 6) The slope, being the change in rise divided by the change in x: (d`7 - d`2) - (d`6 - d`2.5) = 5 - 3.5 = 1.429 1.429 y/x confidence assessment: 2
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18:21:49 Your straight line should pass above the first and third points and beneath the second. If this is not the case the line can be modified so that it comes closer on the average to all three points. The best possible straight line passes through the y-axis near y = 2. The x = 2 point on the best possible line has a y coordinate of about 3, and the x = 7 point has a y coordinate of about 7. So the best possible straight line contains points with approximate coordinate (2,3) and (5,7). The slope between these two points is rise/run = (7 - 3)/(5 - 2) = 4 / 5 = .8. Note that the actual slope and y intercept of the true best-fit line, to 3 significant figures, are .763 and 1.79. So the equation of the line is .763 x + 1.79
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RESPONSE --> My figures seem to be off rather significantly from yours. Not quite sure how. I've double-checked all my plotting and math, Maybe my graph is not quite to scale good enough. I understand how this works though. self critique assessment: 1
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18:47:55 `q002. Plug coordinates of the x = 2 and x = 7 points into the form y = m x + b to obtain two simultaneous linear equations. Give your two equations. Then solve the equations for m and b and substitute these values into the form y = m x + b. What equation do you get?
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RESPONSE --> y = 2m + 3 y = 7m + 7 y = 2m + 3 -3 = 2m -1.5 = m y = 2(-1.5) + 3 y = -3 + 3 y = 0 y = 7m + 7 -7 = 7m -1 = m y = 7(-1) + 7 y = -7 + 7 y = 0 confidence assessment: 1
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18:53:33 Plugging the coordinates (2,3) and (7, 6) into the form y = m x + b we obtain the equations 3 = 2 * m + b 5 = 7 * m + b. Subtracting the first equation from the second will eliminate b. We get 4 = 5 * m. Dividing by 5 we get m = 4/5 = .8. Plugging m = .8 into the first equation we get 3 = 2 * .8 + b, so 3 = 1.6 + b and b = 3 - 1.6 = 1.4. Now the equation y = m x + b becomes y = .8 x + 1.4. Note that the actual best-fit line is y = .763 x + 1.79, accurate to three significant figures.
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RESPONSE --> I had forgotten the method of subtracting one equation from the other. However, I don't understand how doing so yields a result of 4 = 5m self critique assessment: 1
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18:56:43 `q003. Using the equation y = .8 x + 1.4, find the coordinates of the x = 1, 3, and 6 points on the graph of the equation.
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RESPONSE --> Plugging the x=1 point into the equation: y = .8*1 + 1.4 y = .8 + 1.4 y = 2.2 This yields a coordinate of (1 , 2.2) Plugging the x=3 point into the equation: y = .8*3 + 1.4 y = 2.4 + 1.4 y = 3.8 This yields a coordinate of (3 , 3.8) Plugging the x=6 point into the equation: y = .8*6 + 1.4 y = 4.8 + 1.4 y = 6.2 This yields a coordinate of (6 , 6.2) confidence assessment: 2
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18:56:53 Evaluating y =.8 x + 1.4 at x = 1, 3, and 6 we obtain y values 2.2, 3.8, and 6.2.
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RESPONSE --> I understand. self critique assessment: 2
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18:58:42 `q004. The equation y = .8 x + 1.4 gives you points (1, 2.2), (3, 3.8), and (6,6.2). How close, on the average, do these points come to the original data points (1,2), (3, 5), and (6, 6)?
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RESPONSE --> All x-values are identical and the y-values typically vary by a few tenths of a unit, except the (3 , 5) point whose y-value is a whole 1.2 unit less. confidence assessment: 2
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18:59:13 (1, 2.2) differs from (1, 2) by the .2 unit difference between y = 2 and y = 2.2. (3, 3.8) differs from (3, 5) by the 1.2 unit difference between y = 5 and y = 3.8. (6, 6.2) differs from (6, 6) by the .2 unit difference between y = 6 and y = 6.2. {}The average discrepancy is the average of the three discrepancies: ave discrepancy = ( .2 + 1.2 + .2 ) / 3 = .53.
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RESPONSE --> I should've gone into closer detail, but I understand. self critique assessment: 1
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19:04:56 `q005. Using the best-fit equation y =.76 x + 1.79, with the numbers accurate to the nearest .01, how close do the predicted points corresponding to x = 1, 3, and 6 come to the original data points (1,2), (3, 5), and (6, 6)?
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RESPONSE --> y = .76*1 + 1.79 y = .76 + 1.79 y = 2.55 y = .76*3 + 1.79 y = 2.28 + 1.79 y = 4.07 y - .76*6 + 1.79 y = 4.56 + 1.79 y = 6.35 The data corresponding to the point (1,2) is .35 higher. The data corresponding to the point (3,5) is .27 higher. The data corresponding to the point (6,6) is .15 higher. confidence assessment: 2
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19:05:18 Evaluating y =.76 x + 1.79 at x = 1, 3 and 6 we obtain y values 2.55, 4.07 and 6.35. This gives us the points (1,2.55), (3,4.07) and (6, 6.35). These points lie at distances of .55, .93, and .35 from the original data points. The average distance is (.55 + .93 + .35) / 3 = .58 from the points.
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RESPONSE --> I understand. self critique assessment: 2
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19:11:12 `q006. The average distance of the best-fit line to the data points appears to greater than the average distance of the line we obtain by an estimate. In fact, the best-fit line doesn't really minimize the average distance but rather the square of the average distance. The distances for the best-fit model are .55, .93 and .35, while the average distances for our first model are .2, 1.2 and .2. Verify that the average of the square distances is indeed less for the best-fit model.
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RESPONSE --> Average of the square distances: (.55 + .93 + .35) / 3 = .58 Average distance for the first model: (.2 + 1.2 + .2) / 3 = .53 It appears to me that the average of the square distances is more for the best-fit model. confidence assessment: 2
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19:12:03 The square distances for the best-fit model are .55^2 = .30, .93^2 = .87 and .35^2 = .12. The average of these square distances is (.30 + .87 + .12) / 3 = .43. The squared distances for the first model are .2^2 = .04, 1.2^2 = 1.44 and .2^2 = .04, so the average of the square distances for this model is (.04 + 1.44 + .04) / 3 = .51. Thus the best-fit model does give the better result. We won't go into the reasons here why it is desirable to minimize the square of the distance rather than the distance. When doing eyeball estimates, you don't really need to take this subtlety into account. You can simply try to get is close is possible, on the average, to the points.
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RESPONSE --> I didn't understand the question and didn't realize we were supposed to square all the distances. self critique assessment: 1
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19:32:01 `q007. If the original data points (1,2), (3, 5), and (6, 6) represent the selling price in dollars of a bag of widgets vs. the number of widgets in the bag, then how much is paid for a bag of 3 widgets? How much would you estimate a bag of 7 widgets would cost?
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RESPONSE --> The price for a bag of 3 widgets: 3 = .8x + 1.4 1.6 = .8x 2 = x $2.00 for a bag of 3 widgets The price for a bag of 7 widgets: 7 = .8x + 1.4 5.6 = .8x 7 = x $7.00 for a bag of 7 widgets. confidence assessment: 2
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19:34:06 If we use the best-fit function y =.76 x + 1.79, noting that y represents the cost and x the number of widgets, then the cost of 3 widgets is y = .76 * 3 + 1.79 = 4.05, representing cost of $4.05. The cost of 7 widgets would be y = .76 * 7 + 1.79 = 7.11. The cost of 7 widgets would be $7.11.
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RESPONSE --> I accidentally used the y = .8x + 1.4 function instead of the more precise y = .763x + 1.79 best-fit function. I understand how this works though. self critique assessment: 1
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19:36:23 `q008. According to the function y = .8 x + 1.4, how much will a bag of 7 widgets cost? How many widgets would you expect to get for $10?
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RESPONSE --> The price for a bag of 7 widgets: y = .8*7 + 1.4 y = 5.6 + 1.4 y = 7 $7.00 for a bag of 7 widgets The number of widgets $10 will buy: 10 = .8x = 1.4 8.6 = .8x 10.75 = x $10.00 for a bag of 10.75 widgets. confidence assessment: 2
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19:36:38 Using the model we obtained, y = .8 x + 1.4, we note that the cost is represented by y and the number of widgets by acts. Thus we can find cost of 7 widgets by letting x = 7: cost = y = .8 * 7 + 1.4 = 7. To find the number of widgets you can get for $10, let y = 10. Then the equation becomes 10 = .8 x + 1.4. We easily solve this equation by subtracting 1.4 from both sides than dividing by .8 to obtain x = 10.75. That is, we can buy 10.75 widgets with $10.
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RESPONSE --> I understand. self critique assessment: 2
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?????v????T??D???assignment #007 007. `query 7 Precalculus I 10-02-2007
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20:19:38 query predictions Sketch your graph representing the predicted height of the low end vs. the weight on the spring.
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RESPONSE --> I think the more weight that's added to the spring the more it will stretch out. This prediction is supported by the fact that the string didn't even begin stretching until 2 kg of weight is hung from it. confidence assessment: 3
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20:19:58 STUDENT RESPONSE: I predict that the spring will stretch at a greater rate as the weight is added. In my experience, springs lose their strength as they are stretched, and will not go back to their original shape. INSTRUCTOR COMMENT: ** Within their range of elasticity the graph is very nearly linear. If stretched too far the spring will lose its permanent elastic properties and will then deviate from linearity **
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RESPONSE --> I understand. self critique assessment: 2
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20:35:43 comment on how the actual graph of the data compared with your prediction
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RESPONSE --> From my first point of reference (3 , 44.7) the graph appeared to be roughly linear. There is no apparent slope to it as I'd predicted. confidence assessment: 2
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20:36:15 ** If you predicted a linear graph then did the actual graph confirm this? If you predicted a curvature did the actual graph confirm this? **
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RESPONSE --> I predicted a curvature and the graph confirmed linear. self critique assessment: 2
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20:43:37 query linked outline discuss your experience with the Linked Outline. Did you find it helpful?
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RESPONSE --> I studied all the statements on it, and plan on using this as a primary study tool for future tests. I find it extremely helpful. confidence assessment: 2
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20:43:45 ** Many students find the Linked Outline very helpful. **
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RESPONSE --> I understand. self critique assessment: 2
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20:45:00 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Aside from being surprised that the spring would NOT stretch more as more weight was added to it, I found it helpful to see some of the concepts I've been learning in the past few weeks be applied to a real-world situation. confidence assessment: 2
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